Let $G$ be a group and let $H$ be its automorphism group $H = Aut(G)$. However, suppose that we did not have knowledge of $G$. Is there a way to find the outer automorphism group of $G$, $Out(G)$ just by using $H$?


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    $\begingroup$ $Aut(G)/Inn(G)$? $\endgroup$ – Mikasa Mar 17 '13 at 6:17

No. For example: If $n>6$, then $S_n\simeq Aut(S_n)\simeq Aut(A_n)$, so the possibilities $Out(G)=Out(S_n)=1$ and $Out(G)=Out(A_n)=C_2$ both remain.

  • $\begingroup$ Ok. Thanks. an have a great times. $\endgroup$ – Mikasa Mar 17 '13 at 7:23
  • $\begingroup$ @Babak: Your suggestion of $S_3$ works as well: $Aut(S_3)=Aut(C_2\times C_2)=S_3$ $\endgroup$ – Jyrki Lahtonen Mar 17 '13 at 19:28
  • $\begingroup$ May I add the link. May I ask to take a look at this question? I wanna examine a way in which I can probe the continuity a function at origin. Thanks. $\endgroup$ – Mikasa Mar 18 '13 at 9:17

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